SOLUTION: If x+y<16 but x-y>6, where x and y are integers, what is the largest integral value of xy?

The product of two POSITIVE numbers is greatest when their difference in
absolute value is as small as possible.  Since x-y > 6, the smallest their
difference can be is 7, and since they can both be negative integers, their sum
will always be less than 16, so their POSITIVE product gets larger and larger
without bound.

For instance if -10000 and y=-10007, x+y=-20007 < 16 and x-y=7>6,
then xy=100070000.  And we can get xy larger and larger.

There is no solution because there can be no largest value of xy.

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Perhaps you meant that x and y are both POSITIVE integers, and cannot be
negative?  If so, then there is a solution.

Since to get the largest product, their difference must be as small as
possible, and the smallest their difference can be is 7, then x = y+7,

x+y < 16
Substituting y+7 for x
y+7+y < 16
 2y+7 < 16
   2y < 9
    y < 4.5 and since y is an integer, the largest y can be is 4, and
the largest x can be is 4+7 or 11.  So the largest possible product xy
is (4)(11) = 44.  That is the solution if x and y are POSITIVE integers.

Edwin